0.2 kg of water vanar al `100^(@)C` is admitted into a mixture consisting of 5 kg of water and 3 kg of ice. What will happen? Neglect radiative losses.

Explanation

The heat liberated in the processes of condensation of water vapour and of cooling of water down to the melting point of ice is `Q=m_(3) (L+c Deltat)=5.2xx10^(5)J`. This is not enough to melt all the ice. For this the required heat is `Q_(m)=m_(2)lambda=10^(6)J`. Therefore the ice will only partially melt.